Question

Show that the tangent to $y^2=4ax$ at $(a,2a)$ touches the parabola $x^2=4ay$.

Answer 1

We have,

Equation of parabola ( 1),

$y^2=4ax$

On differentiating and we get,

$2y\frac{dy}{dx}=4a$

$\Rightarrow \frac{dy}{dx}=\frac{2a}{y}$

At the point (a,2a)

$\frac{dy}{dx}=\left(\frac{2a}{2a}\right)=1$

Now, Equation of parabola (2),

$x^2=4ay$

On differentiating and we get,

$2x=4a\frac{d{y}_1}{dx}$

$\frac{d{y}_1}{dx}=\frac{x}{2a}$

At the point $(a,2a)$

$\frac{d{y}_1}{dx}=\left(\frac{a}{2a}\right)=\frac{1}{2}$

Then,

$\frac{dy}{dx}\ne \frac{d{y}_1}{dx}$

Hence, it is not common tangent.